Assymptotic approximation of $\int x^n \log xdx$ Hi I'm following an Analysis course and one of the exercises is to prove that:


*

*$$\int_{1/2}^1 x^n \log x dx \approx -1/n^2 $$

*$$\left|\int_0^{1/2} x^n \log x dx\right| = O\left((1/2)^n\right)$$


At first I thought to solve this by the laplace method and write the integrand as a power of $e$ and then simplified by Taylor series. 
Is there someone who can help me so that I will understand how to solve this asymptotic approximation.
Many thanks!!
 A: The integral can be evaluated in closed form.  Integrating by parts, we have
$$\begin{align}
\int_a^b x^n \log x\,dx&=\left.\frac{x^{n+1}\log x}{n+1}\right|_a^b-\frac{1}{n+1}\int_a^bx^n\,dx\\\\
&=\left.\frac{x^{n+1}\log x}{n+1}\right|_a^b-\left. \frac{x^{n+1}}{(n+1)^2}\right|_a^b\\\\
&=\frac{b^{n+1}\log b-a^{n+1}\log a}{n+1}-\frac{b^{n+1}-a^{n+1}}{(n+1)^2}
\end{align}$$
Setting $a=1/2$ and $b=1$ for Part $1$ and setting $a=0$ and $b=1/2$ for Part $2$ yield the desired results.

Suppose one is not permitted to use integration by parts at this stage.  Then for the integral $\int_a^b x^n \log x\,dx=\int_a^b e^{n\log x}\,\log x\,dx$, we can write (in the spirit of Laplace's Method)
$$\log x \sim  \log b +\frac1b(x-b)$$
Therefore, we have
$$\begin{align}
\int_a^b e^{n\log x}\,\log x\,dx &\sim \int_a^b e^{n\log b+\frac{1}{b}(x-b)}\left(\log b+\frac{n}{b}(x-b)\right)\,dx\\\\
&=\frac{b^{n+1}\log b}{n}\left(1-e^{-n(b-a)/b}\right)-\frac{b^{n+1}}{n^2}\int_0^{n(b-a)/b}\,te^{-t}\,dt\tag 1
\end{align}$$
whereupon applying the specific value for $a$ and $b$, we can deduce the asymptotic limiting behavior without integrating by parts the integral in $(1)$.
A: Hint: If $~F(n)=\displaystyle\int x^n~dx,~$ then $F'(n)=\displaystyle\int x^n\ln x~dx$.
